If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)]


If [2sinα / (1+cosα+sinα)] = y, then prove that [(1– cosα+sinα) / (1+sinα)] is also equal to y.

$\left[\right.$ Hint : Express $\left.\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\right]$


According to the question,

y =2sinα /(1+cosα+sinα)

Multiplying numerator and denominator by (1 – cos α + sin α),

We get,

$\Rightarrow y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha} \times \frac{1-\cos \alpha+\sin \alpha}{1-\cos \alpha+\sin \alpha}$

$=\frac{2 \sin \alpha}{(1+\sin \alpha)+\cos \alpha} \times \frac{(1+\sin \alpha)-\cos \alpha}{(1+\sin \alpha)-\cos \alpha}$

Using $(a+b)(a-b)=a^{2}-b^{2}$, we get:

$=\frac{2 \sin \alpha\{(1+\sin \alpha)-\cos \alpha\}}{(1+\sin \alpha)^{2}-\cos ^{2} \alpha}$

$=\frac{2 \sin \alpha(1+\sin \alpha)-2 \sin \alpha \cos \alpha}{1+\sin ^{2} \alpha+2 \sin \alpha-\cos ^{2} \alpha}$

Since, $1-\cos ^{2} \alpha=\sin ^{2} \alpha$

$\therefore y=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{\sin ^{2} \alpha+2 \sin \alpha+\sin ^{2} \alpha}$

$=\frac{2 \sin \alpha(1+\sin \alpha-\cos \alpha)}{2 \sin \alpha(1+\sin \alpha)}$

$\Rightarrow y=\frac{(1+\sin \alpha-\cos \alpha)}{(1+\sin \alpha)}$

$\Rightarrow y=\frac{(1-\cos \alpha+\sin \alpha)}{(1+\sin \alpha)}$

Hence Proved

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